If a farmer stocked his pond with 100 fish in the year 2000 and their population grows at a fixed rate of 10% how many fish will he have in the year 2035? hint: use the rule of
70!

By 2035 he will have 2810 fishes.

Solution:

Given, if a farmer stocked his pond with 100 fish in the year 2000 and their population grows at a fixed rate of a year

We have to find number of fish he will have in the year 2035. Hint: Use the Rule of 70!  

Let the amount of fish in the tank be x. initially x = 100

If x grows , then you have the original x plus of x.  

Changing from percentage to decimal form (which is almost always necessary and is necessary this time) so the growth is and what was kept was 1x.  

When you add up what you keep (the original principal) and the interest that you gain you have  

If you do this for two years it is and after n many years you will have

In this problem, n = 35 and x = 100  

So, the answer is which rounds down to 2,810 fish.

associative property of multiplication

answer:

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

explanation:

service only with 20% tip  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

in the opening of the question the word 'service' is associated only with the pretax value.

the closing part of the question asks: how much did she pay for the  

service

−−−−−−  

so technically the answer is  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

however, the intention of the question provider may have been that your answer is to be  

$

87.12

(

1

+

8

100

)

=

$

94.09

to 2 decimal places.

'

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

tony b

mar 21, 2017

answer:

trying to explain it another way

explanation:

service charge  

.

→

$

72.60

20% tip  

20

100

×

$

72.60

=

$

14.52

−−−−−−  

←

add

$

87.12

tax  

8

100

×

$

87.12

=

$

6.97

−−−−−  

←

add

$

94.09

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